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2 years ago

A suspended object A is attracted to a charged object B, can one conclude that A is charged? Explain

Physics

1 answer:

irinina [24]2 years ago

5 0

Explanation:

In my view, when the Object A is attracted to a Charged object B. Object B should be Negatively or Positively charged. So Object B should be the Opposite charged according to the Object B

Example =

If Object B is Negatively Charged, the Object A should be Positively Charged

If the Object B is Positively Charged, the Object A should be Negatively Charged

Sometimes it can Mix as a Neutral as well

Hope this Helps

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Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom

Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

$B = B_o(-\hat k)$

now the velocity is from bottom to top

so we have

$v = v_o \hat j$

now the force on the moving charge is given as

$\vec F = q(\vec v \times \vec B)$

now we have

$\vec F = (-e)(v_o \hat j \times B_o(-\hat k))$

$\vec F = e v_o B \hat i$

so force will be towards Right

7 0

2 years ago

a microwave operates at a frequency of 2400 MHZ. the height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. as

alexira [117]

Complete question is;

A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly

downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.

Answer:

Power ≈ 600,000 W

Explanation:

We are given;

Frequency; f = 2400 Hz

height of the oven cavity; h = 25 cm = 0.25 m

base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²

total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J

We want to find the power output and we know that formula for power is;

P = workdone/time taken

Formula for time here is;

t = h/c

Where c is speed of light = 3 × 10^(8) m/s

Thus;

t = 0.25/(3 × 10^(8))

t = 8.333 × 10^(-10) s

Thus;

Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))

Power ≈ 600,000 W

3 0

2 years ago

An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.

ivann1987 [24]

<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters

</span>

5 0

2 years ago

A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus

babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

$I = m\cdot (\vec{v}_{2} - \vec{v_{1}})$ (1)

Where:

$I$ - Impulse, in kilogram-meters per second.

$m$ - Mass, in kilograms.

$\vec{v_{1}}$ - Initial velocity of the hockey park, in meters per second.

$\vec{v_{2}}$ - Final velocity of the hockey park, in meters per second.

If we know that $m = 0.2\,kg$ , $\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]$ and $\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]$ , then the impulse applied by the stick to the park is approximately: